ICAR JRF Plant Science Practice Series Memory Based 2024 (Module 1) (41 – 80 MCQ) 

 

Question 41

Options:

1. Testa
2. Micropyle
3. Tegmen
4. Epithelium

Correct Answer: 1

 

Explanation:

• The specific question text is missing, but the answer key indicates Testa→ Option 1.
• This likely asks: Which is the outermost seed coat?
• Testa– The outer seed coat, derived from the outer integument of the ovule.
• Tegmen– Inner seed coat, derived from the inner integument.
• Micropyle– Small opening in the seed coat.
• Epithelium– A layer of cells lining a surface.
• Therefore, the correct answer is Testa→ Option 1.

 

 

Question 42 

The organic compound glucosinolates are present in members of the family:

1. Caryophyllaceae
2. Chenopodiaceae
3. Euphorbiaceae
4. Brassicaceae

Correct Answer: 4

 

Explanation:

• Glucosinolates– Sulfur-containing secondary metabolites found primarily in the family Brassicaceae (Cruciferae).
• They are responsible for the pungent taste and aroma of mustard, cabbage, broccoli, cauliflower, and radish.
• When hydrolyzed by the enzyme myrosinase, they produce isothiocyanates (e.g., allyl isothiocyanate in mustard).
• Caryophyllaceae– Carnation family; produces saponins, not glucosinolates.
• Chenopodiaceae– Beet, spinach family; produces betalains and oxalates.
• Euphorbiaceae– Castor, rubber family; produces latex, cyanogenic glycosides.
• Therefore, the correct answer is Brassicaceae→ Option 4.

 

 

Question 43 

In a self-pollinated crop variety, if a mechanical mixture was caused during threshing, then which breeding method would you adopt to purify the variety?

1. Recurrent selection
2. Bulk selection
3. Mass selection
4. Pure line selection

Correct Answer: 4

 

Explanation:

• Pure line selection– The method used to purify a self-pollinated crop variety that has become mixed (mechanical mixture). Individual plants are selected based on desirable traits, and their progeny are evaluated and multiplied.
• Mass selection– Selecting and bulking seeds from many desirable plants; used for population improvement but not for purifying a pure line variety.
• Bulk selection– Growing a population in bulk without individual plant selection; used for natural selection and adaptation.
• Recurrent selection– Used for population improvement in cross-pollinated crops.
• Since the variety is self-pollinated and a pure line is desired, pure line selectionis the most appropriate → Option 4.

 

 

Question 44 

Options:

1. Cauliflower mosaic virus
2. Banana bunchy top virus
3. Bacilliform DNA virus
4. Rice dwarf virus

Correct Answer: 1

 

Explanation:

• The specific question text is missing, but the answer key indicates Cauliflower mosaic virus→ Option 1.
• Cauliflower mosaic virus (CaMV)– A plant virus with double-stranded DNA (dsDNA) genome, belonging to the family Caulimoviridae.
• Other options are either ssDNA or dsDNA viruses, but CaMV is the classic example.
• Therefore, the correct answer is Cauliflower mosaic virus→ Option 1.

 

 

Question 45 

The three-nucleotide combination, adenine-uracil-guanine codes for:

1. Methionine
2. Leucine
3. Alanine
4. Guanine

Correct Answer: 1

 

Explanation:

• AUG is the start codonin the genetic code.
• It codes for methionine(Met) in eukaryotes and formylmethionine (fMet) in prokaryotes.
• AUG is also the initiation codon for protein synthesis.
• Therefore, the correct answer is Methionine→ Option 1.

 

 

Question 46 

The type of germination in which seeds germinate by keeping cotyledons below the soil surface due to rapid elongation of epicotyl, is referred to as:

1. Epigeal
2. Hypogeal
3. Hypo-epigeal
4. Epi-hypogeal

Correct Answer: 2

 

Explanation:

• Hypogeal germination– The cotyledons remain below the soil surface because the epicotyl elongates, pushing the plumule upward. The cotyledons do not become photosynthetic.
• Examples:Pea, maize, wheat, rice, gram.
• Epigeal germination– The hypocotyl elongates, pushing the cotyledons above the soil (e.g., bean, castor, onion).
• Hypo-epigeal / Epi-hypogeal– Not standard terms.
• Therefore, the correct answer is Hypogeal→ Option 2.

 

 

Question 47 

Options:

1. Lactobacillus plantarum
2. Clostridium butyricum
3. Bacillus subtilis
4. Bacillus megaterium

Correct Answer: 2

 

Explanation:

• The specific question text is missing, but the answer key indicates Clostridium butyricum→ Option 2.
• Clostridium butyricumis a Gram-positive, spore-forming, anaerobic bacterium that produces butyric acid.
• It is used in the production of butanoland as a probiotic.
• Therefore, the correct answer is Clostridium butyricum→ Option 2.

 

 

Question 48 

The isolation distance to be maintained for retaining the genetic purity of parental lines of maize hybrids while producing foundation seed is:

1. 400 meters
2. 1000 meters
3. 100 meters
4. 200 meters

Correct Answer: 4

 

Explanation:

• Isolation distancefor seed production of parental lines of maize hybrids for foundation seed is 200 meters as per Indian Minimum Seed Certification Standards.
• For certified seedof maize hybrids, isolation distance is 200-400 m depending on the class.
• For open-pollinated varietiesof maize, isolation distance is 400 m.
• Option 1 (400 m)– For OPVs or certified seeds of some cross-pollinated crops.
• Option 2 (1000 m)– For foundation seed of certain cross-pollinated crops like sunflower.
• Option 3 (100 m)– For self-pollinated crops.
• Therefore, the correct answer is 200 meters→ Option 4.

 

 

Question 49 

In the seeds of most flowering plants, a triploid albuminous tissue is found which is botanically described as:

1. Seed coat
2. Cotyledon
3. Endosperm
4. Embryo

Correct Answer: 3

 

Explanation:

• Endosperm– The triploid (3n) albuminous tissue in angiosperm seeds formed by double fertilization (fusion of one sperm with two polar nuclei).
• It serves as a nutritive tissue for the developing embryo.
• Seed coat– Derived from integuments (diploid).
• Cotyledon– Part of the embryo (diploid).
• Embryo– Diploid (zygote).
• Therefore, the correct answer is Endosperm→ Option 3.

 

Question 50 

Options:

1. Griffith
2. Tatum
3. Lederberg
4. Zinder & Lederberg

Correct Answer: 4

 

Explanation:

• The specific question text is missing, but the answer key indicates Zinder & Lederberg→ Option 4.
• This likely asks: Who discovered transduction?
• Zinder and Lederberg (1952)– Discovered transduction in Salmonella typhimurium using phage P22.
• Griffith– Discovered transformation (1928).
• Tatum– Worked on one gene-one enzyme hypothesis (with Beadle).
• Lederberg– Worked on conjugation and transduction.
• Therefore, the correct answer is Zinder & Lederberg→ Option 4.

 

Question 51 

Etiology is the study of:

1. Cause of disease
2. Symptoms of disease
3. Predisposing factors for disease
4. Management of disease

Correct Answer: 1

 

Explanation:

• Etiology– The branch of science that deals with the cause or origin of disease.
• Symptoms– Studied under symptomatology.
• Predisposing factors– Part of epidemiology.
• Management– Part of disease control.
• Therefore, the correct answer is Cause of disease→ Option 1.

 

Question 52

Which of the following is present in a prokaryotic cell?

1. Nucleoid
2. Nuclear membrane
3. Mitochondria
4. 80 S ribosomes

Correct Answer: 1

 

Explanation:

• Nucleoid– The region in a prokaryotic cell where the genetic material (circular DNA) is located, not enclosed by a nuclear membrane.
• Nuclear membrane– Absent in prokaryotes.
• Mitochondria– Absent in prokaryotes (respiration occurs on the plasma membrane).
• 80S ribosomes– Present in eukaryotes; prokaryotes have 70S ribosomes.
• Therefore, the correct answer is Nucleoid→ Option 1.

 

Question 53 

Options:

1. White rust
2. Yellow rust
3. Black rust
4. Red rust

Correct Answer: 2

 

Explanation:

• The specific question text is missing, but the answer key indicates Yellow rust→ Option 2.
• This likely asks: Which rust is caused by Puccinia striiformis?
• Yellow rust(stripe rust) of wheat – caused by Puccinia striiformis sp. tritici.
• Therefore, the correct answer is Yellow rust→ Option 2.

 

Question 54 

Cell wall of a bacteria is made up of:

1. Cellulose
2. Chitin
3. Peptidoglycan
4. Dextran

Correct Answer: 3

 

Explanation:

• Peptidoglycan(murein) – A polymer consisting of sugars (N-acetylglucosamine and N-acetylmuramic acid) and amino acids. It provides structural strength to bacterial cell walls.
• Cellulose– Found in plant cell walls.
• Chitin– Found in fungal cell walls and arthropod exoskeletons.
• Dextran– A polysaccharide produced by some bacteria; not a cell wall component.
• Therefore, the correct answer is Peptidoglycan→ Option 3.

 

Question 55 

Which of the following is NOT a gram +ve bacterial plant pathogenic genus?

1. Rathayibacter
2. Streptomyces
3. Xanthomonas
4. Clavibacter

Correct Answer: 3

 

Explanation:

• Xanthomonas– Gram-negative bacterium (plant pathogen causing leaf spots, blights, cankers).
• Rathayibacter– Gram-positive (causes bacterial wilt in grasses).
• Streptomyces– Gram-positive (causes common scab of potato).
• Clavibacter– Gram-positive (causes bacterial ring rot of potato and wilt of tomato).
• Therefore, the genus NOTgram-positive is Xanthomonas → Option 3.

 

Question 56 

Options:

1. Peritrichous
2. Monotrichous
3. Lophotrichous
4. Atrichous

Correct Answer: 1

 

Explanation:

• The specific question text is missing, but the answer key indicates Peritrichous→ Option 1.
• Peritrichous– Flagella distributed all over the bacterial cell surface (e.g., Escherichia coli, Salmonella).
• Therefore, the correct answer is Peritrichous→ Option 1.

 

Question 57 

When a disease becomes prevalent throughout the country, continent or the world in a severe form, it is referred to as:

1. Epidemic
2. Sporadic
3. Endemic
4. Pandemic

Correct Answer: 4

 

Explanation:

• Pandemic– A disease that spreads across a large geographic area (multiple countries, continents, or worldwide) affecting a high proportion of the population.
• Epidemic– Sudden increase in disease cases within a specific region or population.
• Sporadic– Occasional, scattered cases.
• Endemic– Constantly present at a baseline level in a specific area.
• Therefore, the correct answer is Pandemic→ Option 4.

 

Question 58 

ICAR- National Bureau of Agriculturally Important Microorganisms is located at:

1. Ludhiana
2. Shimla
3. Mau
4. Bangalore

Correct Answer: 3

 

Explanation:

• ICAR-National Bureau of Agriculturally Important Microorganisms (NBAIM)is located at Mau Nath Bhanjan, Uttar Pradesh.
• Ludhiana– ICAR-Indian Institute of Maize Research.
• Shimla– ICAR-Central Potato Research Institute.
• Bangalore– ICAR-National Bureau of Agricultural Insect Resources (NBAIR).
• Therefore, the correct answer is Mau→ Option 3.

 

Question 59 

Options:

1. Jute
2. Cotton
3. Sisal
4. Manilla hemp

Correct Answer: 2

 

Explanation:

• The specific question text is missing, but the answer key indicates Cotton→ Option 2.
• This likely asks: Which is a surface fibre?
• Cotton– Surface fibre (seed hair/trichome).
• Jute, Sisal, Manilla hemp– Bast fibres (stem fibres).
• Therefore, the correct answer is Cotton→ Option 2.

 

Question 60 

Which of the following is a pleomorphic fungus?

1. Sclerospora
2. Erysiphe
3. Ustilago
4. Puccinia

Correct Answer: 4

 

Explanation:

• Pleomorphic fungi– Fungi that produce two or more different types of spores during their life cycle (have both asexual and sexual stages; also called holomorphic fungi).
• Puccinia– A rust fungus that produces five different spore stages (pycniospores, aeciospores, urediniospores, teliospores, basidiospores). It is a classic example of a pleomorphic fungus.
• Sclerospora– Downy mildew fungus; produces zoospores and oospores.
• Erysiphe– Powdery mildew; produces conidia and cleistothecia.
• Ustilago– Smut fungus; produces teliospores and basidiospores.
• Therefore, the correct answer is Puccinia→ Option 4.

 

Question 61 

The fungi in which parasitic habit is weakly developed is called:

1. Perthotrophs
2. Facultative parasites
3. Necrotrophs
4. Facultative saprophytes

Correct Answer: 2

 

Explanation:

• Facultative parasites– Organisms that are normally saprophytic but can become parasitic under certain conditions. Their parasitic habit is weakly developed (they prefer dead organic matter but can infect living hosts when opportunity arises).
• Perthotrophs– Not a standard term.
• Necrotrophs– Kill host tissue and feed on dead cells (strong parasitic habit).
• Facultative saprophytes– Normally parasitic but can survive as saprophytes (e.g., Fusarium).
• Therefore, the correct answer is Facultative parasites→ Option 2.

 

Question 62 

Options:

1. Corticous
2. Epixylic
3. Coprophilous
4. Xylophilous

Correct Answer: 3

 

Explanation:

• The specific question text is missing, but the answer key indicates Coprophilous→ Option 3.
• Coprophilous– Fungi that grow on dung (dung-inhabiting fungi).
• Therefore, the correct answer is Coprophilous→ Option 3.

 

Question 63 

Sphacelia phase is the characteristic of:

1. Penicillium
2. Aspergillus
3. Claviceps
4. Curvularia

Correct Answer: 3

 

Explanation:

• Sphacelia phase– The asexual (conidial) stage of the ergot fungus Claviceps purpurea.
• In this phase, the fungus produces conidiain a sugary liquid called honeydew, which is spread by insects.
• Later, the sclerotium (ergot body) replaces the sphacelia.
• Penicillium, Aspergillus, Curvularia– Do not have a sphacelia phase.
• Therefore, the correct answer is Claviceps→ Option 3.

 

Question 64 

Which of the following belongs to the imperfect fungi?

1. Puccinia
2. Phytophthora
3. Ustilago
4. Rhizoctonia

Correct Answer: 4

 

Explanation:

• Imperfect fungi– Fungi classified under Deuteromycota (no known sexual stage).
• Rhizoctonia– A deuteromycete (no sexual stage known, though teleomorph is Thanatephorus in Basidiomycota). It is a soilborne pathogen causing damping-off, root rot, and sheath blight.
• Puccinia– Basidiomycete (rust fungus).
• Phytophthora– Oomycete (not a true fungus; belongs to Chromista).
• Ustilago– Basidiomycete (smut fungus).
• Therefore, the correct answer is Rhizoctonia→ Option 4.

 

 

Question 65 

Options:

1. Moko disease
2. Panama wilt
3. Bunchy top
4. Sigatoka

Correct Answer: 2

 

Explanation:

• The specific question text is missing, but the answer key indicates Panama wilt→ Option 2.
• Panama wilt– Caused by Fusarium oxysporum sp. cubense, a soilborne fungus affecting banana.
• Therefore, the correct answer is Panama wilt→ Option 2.

 

 

Question 66

Which of the following is asexual spores in fungi?

1. Ascospore
2. Basidiospore
3. Oospore
4. Chlamydospore

Correct Answer: 4

 

Explanation:

• Chlamydospore– An asexual spore formed by the thickening of hyphal cells. It is a thick-walled, resistant spore that can survive adverse conditions.
• Ascospore– Sexual spore produced in an ascus (Ascomycota).
• Basidiospore– Sexual spore produced on a basidium (Basidiomycota).
• Oospore– Sexual spore produced by Oomycetes.
• Therefore, the correct answer is Chlamydospore→ Option 4.

 

 

Question 67 

Sclerotinia sclerotiorum produces ascocarp named:

1. Apothecium
2. Pseudothecium
3. Cleistothecium
4. Perithecium

Correct Answer: 1

 

Explanation:

• Sclerotinia sclerotiorum– A plant pathogenic fungus (causes white mold / cottony rot).
• It produces an apothecium(cup-shaped ascocarp) as its sexual stage (teleomorph).
• Apothecium– An open, cup-shaped ascocarp where asci are exposed at maturity.
• Pseudothecium– A flask-shaped ascocarp with a pore (ostiole) – found in some Dothideomycetes.
• Cleistothecium– Closed, spherical ascocarp – found in powdery mildews.
• Perithecium– Flask-shaped ascocarp with an ostiole – found in some ascomycetes.
• Therefore, the correct answer is Apothecium→ Option 1.

 

 

Question 68

In terms of ploidy, trisomy is a form of:

1. Hypoploidy
2. Hyperploidy
3. Polyploidy
4. Diploidy

Correct Answer: 2

 

Explanation:

• Trisomy– The presence of an extra chromosome (2n+1). It is a type of hyperploidy (increase in chromosome number beyond the diploid set in a specific chromosome).
• Hypoploidy– Loss of a chromosome (2n-1, monosomy; 2n-2, nullisomy).
• Polyploidy– Multiplication of entire genome (triploid, tetraploid, etc.).
• Diploidy– Normal chromosome number (2n).
• Therefore, trisomy is a form of hyperploidy→ Option 2.

 

 

Question 69 

The Ill-stage of Puccinia graminis tritici produces:

1. Teletospores
2. Uredospores
3. Aecispores
4. Pycnospores

Correct Answer: 3

 

Explanation:

• Puccinia graminis sp. tritici(black stem rust of wheat) produces five spore stages on two hosts (heteroecious).
• Stage I (Spermogonia/Pycnidia)– Produces pycniospores (spermatia) on barberry.
• Stage II (Aecia)– Produces aeciospores (aecispores) on barberry.
• Stage III (Uredinia)– Produces urediniospores on wheat.
• Stage IV (Telia)– Produces teliospores on wheat.
• Stage V (Basidia)– Produces basidiospores on germinated teliospores.
• Ill-stagerefers to Stage II (Aecia) , which produces aecispores → Option 3.

 

 

Question 70 

Which of the following bacterium is used as a biocontrol agent for the control of Meloidogyne species (root-knot nematodes)?

1. Agrobacterium
2. Rhizobium
3. Pasteuria
4. Erwinia

Correct Answer: 3

 

Explanation:

• Pasteuria– A Gram-positive, endospore-forming, obligate parasitic bacterium used as a biocontrol agent against plant-parasitic nematodes, including Meloidogyne (root-knot nematodes), Heterodera, and Globodera.
• Pasteuria penetransattaches to nematode cuticle, penetrates, and kills the nematode.
• Agrobacterium– Causes crown gall; not a biocontrol agent for nematodes.
• Rhizobium– Symbiotic nitrogen fixer; not a nematode biocontrol agent.
• Erwinia– Causes soft rot; not a nematode biocontrol agent.
• Therefore, the correct answer is Pasteuria→ Option 3.

 

Question 71

Grey mould of chickpea is caused by:

1. Sclerotium rolfsii
2. Ascochyta rabiei
3. Botrytis cinerea
4. Sclerotinia sclerotiorum

Correct Answer: 3

 

Explanation:

• Grey mouldof chickpea is caused by Botrytis cinerea (a necrotrophic fungus). It is also known as Botrytis blight or gray mold.
• Sclerotium rolfsii– Causes collar rot / southern blight.
• Ascochyta rabiei– Causes Ascochyta blight (the most serious disease of chickpea).
• Sclerotinia sclerotiorum– Causes white mold / stem rot.
• Therefore, the correct answer is Botrytis cinerea→ Option 3.

 

 

Question 72 

Which of the following plant diseases are caused by Deuteromycetes?

1. Powdery mildews
2. Ergots
3. Downy mildews
4. Wilts

Correct Answer: 4

 

Explanation:

• Deuteromycetes(Fungi Imperfecti) are fungi with no known sexual stage. Many are soilborne and cause wilt diseases (e.g., Fusarium oxysporum causes Fusarium wilt; Verticillium causes Verticillium wilt).
• Powdery mildews– Caused by Ascomycetes (Erysiphaceae).
• Ergots– Caused by Claviceps purpurea (Ascomycete).
• Downy mildews– Caused by Oomycetes (Chromista), not true fungi.
• Therefore, the correct answer is Wilts→ Option 4.

 

Question 73 

Centre of origin of pigeon pea is:

1. The Mediterranean centre
2. The Chinese centre
3. The South American centre
4. Peninsular India

Correct Answer: 4

 

Explanation:

• Pigeon pea(Cajanus cajan) originated in Peninsular India (Vavilov’s Hindustan Centre of origin).
• It is an important pulse crop grown in tropical and subtropical regions.
• The Mediterranean centre – Origin of wheat, barley, chickpea, lentil.
• The Chinese centre – Origin of soybean, millets, buckwheat.
• The South American centre – Origin of potato, tomato, maize, cassava.
• Therefore, the correct answer is Peninsular India→ Option 4.

 

Question 74 

Minimum population size required for recombination of all F₁ gametes at random is:

1. 2ⁿ
2. 3ⁿ
3. 4ⁿ
4. n²

Correct Answer: 1

 

Explanation:

• In a random mating population derived from F₁ hybrids, the minimum population size required to recover all possible gamete combinations (recombination of all F₁ gametes) is 2ⁿ, where n = number of heterozygous genes.
• This ensures that all possible allele combinations from the F₁ parent are represented in the progeny.
• Therefore, the correct answer is 2ⁿ→ Option 1.

 

 

Question 75 

Options:

1. 30 μm
2. 25 μm
3. 75 μm
4. 00 μm

Correct Answer: 2

 

Explanation:

• The specific question text is missing, but the answer key indicates 25 μm→ Option 2.
• This likely asks for the diameter of a typical bacterial cellor the resolution limit of a light microscope.
• The resolution limit of a light microscope is about 2 μm, so 1.25 μm may refer to bacterial cell size.
• Therefore, the correct answer is 25 μm→ Option 2.

 

 

Question 76 

Multi Parent Advanced Generation Intercross Population (MAGIC) generally requires how many parents in crossing?

1. 100 parents
2. 8 parents
3. 2 parents
4. All members of minicore

Correct Answer: 2

 

Explanation:

• MAGIC (Multi Parent Advanced Generation Intercross)population is derived by intercrossing multiple parents (usually 8 parents) over several generations to create a population with high genetic diversity and fine mapping resolution.
• 8 parentsis the typical number used in many MAGIC populations (e.g., in Arabidopsis, wheat, rice).
• 2 parents– Biparental mapping population (RILs, F₂, BC).
• 100 parents– Too many for practical intercrossing.
• All members of minicore– Not a standard MAGIC design.
• Therefore, the correct answer is 8 parents→ Option 2.

 

 

Question 77 

Ill-stage of Puccinia graminis tritici produces:

1. Teletospores
2. Uredospores
3. Aecispores
4. Pycnospores

Correct Answer: 3

 

Explanation:

• Puccinia graminis sp. tritici(black stem rust of wheat) produces five spore stages on two hosts (heteroecious).
• Stage I (Spermogonia/Pycnidia)– Produces pycniospores (spermatia) on barberry.
• Stage II (Aecia)– Produces aeciospores (aecispores) on barberry.
• Stage III (Uredinia)– Produces urediniospores on wheat.
• Stage IV (Telia)– Produces teliospores on wheat.
• Stage V (Basidia)– Produces basidiospores on germinated teliospores.
• Ill-stagerefers to Stage II (Aecia) , which produces aecispores.
• Therefore, the correct answer is Aecispores→ Option 3.

 

 

Question 78 

Which of the following bacterium is used as a biocontrol agent for the control of Meloidogyne species (root-knot nematodes)?

1. Agrobacterium
2. Rhizobium
3. Pasteuria
4. Erwinia

Correct Answer: 3

 

Explanation:

• Pasteuria– A Gram-positive, endospore-forming, obligate parasitic bacterium that is used as a biocontrol agent against plant-parasitic nematodes, including Meloidogyne (root-knot nematodes), Heterodera, and Globodera.
• Pasteuria penetransattaches to nematode cuticle, penetrates, and kills the nematode.
• Agrobacterium– Causes crown gall; not a biocontrol agent for nematodes.
• Rhizobium– Symbiotic nitrogen fixer; not a nematode biocontrol agent.
• Erwinia– Causes soft rot; not a nematode biocontrol agent.
• Therefore, the correct answer is Pasteuria→ Option 3.

 

 

Question 79 

Grey mould of chickpea is caused by:

1. Sclerotium rolfsii
2. Ascochyta rabiei
3. Botrytis cinerea
4. Sclerotinia sclerotiorum

Correct Answer: 3

 

Explanation:

• Grey mouldof chickpea is caused by Botrytis cinerea (a necrotrophic fungus). It is also known as Botrytis blight or gray mold.
• Sclerotium rolfsii– Causes collar rot / southern blight.
• Ascochyta rabiei– Causes Ascochyta blight (the most serious disease of chickpea).
• Sclerotinia sclerotiorum– Causes white mold / stem rot.
• Therefore, the correct answer is Botrytis cinerea→ Option 3.

 

 

Question 80

In term of ploidy, trisomy is a form of:

1. Hypoploidy
2. Hyperploidy
3. Polyploidy
4. Diploidy

Correct Answer: 2

 

Explanation:

• Trisomy– The presence of an extra chromosome (2n+1). It is a type of hyperploidy (increase in chromosome number beyond the diploid set in a specific chromosome).
• Hypoploidy– Loss of a chromosome (2n-1, monosomy; 2n-2, nullisomy).
• Polyploidy– Multiplication of entire genome (triploid, tetraploid, etc.).
• Diploidy– Normal chromosome number (2n).
• Therefore, trisomy is a form of hyperploidy→ Option 2.